What is the solution for y^{''}-6y^'+5y=3e^{2t},y(0)=2,y^'(0)=3 ?

The solution for y^{''}-6y^'+5y=3e^{2t},y(0)=2,y^'(0)=3 is y=(-5/2+3)e^{5t}+5/2 e^t-e^{2t}

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y^{''}-6y^'+5y=3e^{2t},y(0)=2,y^'(0)=3

y^{^{''}} - 6 y^{^{'}} + 5 y = 3 e^{2 t}, y (0) = 2, y^{^{'}} (0) = 3

y = (- \frac{5}{2} + 3) e^{5 t} + \frac{5}{2} e^{t} - e^{2 t}

Solution steps

y^{''} (t) - 6 y^{'} (t) + 5 y = 3 e^{2 t}

Solve linear ODE:

y = (- \frac{5}{2} + 3) e^{5 t} + \frac{5}{2} e^{t} - e^{2 t}

y = (- \frac{5}{2} + 3) e^{5 t} + \frac{5}{2} e^{t} - e^{2 t}

y^{^{'''}} + 4 y^{^{'}} = e^{x} + 3 sin (2 x) + 1

5 y^{^{''}} + 6 y^{^{'}} + 25 y = 4 cos (2 t)

y^{^{''}} + y^{^{'}} = 3, x^{2} y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} - 2 y^{^{'}} - 7 y = 1 e^{5 t}

9 y^{^{''}} + 4 y^{^{'}} - y = 22

What is the solution for y^{''}-6y^'+5y=3e^{2t},y(0)=2,y^'(0)=3 ?
The solution for y^{''}-6y^'+5y=3e^{2t},y(0)=2,y^'(0)=3 is y=(-5/2+3)e^{5t}+5/2 e^t-e^{2t}